对模板成员的奇怪“未定义引用”

Question

首先，我知道必须在头文件中实现定义，这就是我通常根据this thread所做的

我有几个文件

// Image.h
namespace library
{
    namespace data
    {
        template<class Pixel, class SubPixel, int Layer>
        class Image 
        {
            public:
            Image(int width, int height, Pixel emptyFill);
            Image(int width, int height);
            ~Image();

            Pixel& operator() (int x, int y);
            const Pixel& operator() (int x, int y) const;
            SubPixel& operator() (int x, int y, int subpixel);
            const SubPixel& operator() (int x, int y, int subpixel) const;

            protected:
            int m_width;
            int m_height;

            Pixel* m_pixels;
            void init(Pixel emptyFill);
            bool checkBounds(int x, int y, int layer = 0) const;

            virtual SubPixel& pixelToSubpixel(Pixel& pixel, int layer) const = 0;
        };

        template<class Pixel, class SubPixel, int Layer>
        Image<Pixel, SubPixel, Layer>::Image(int width, int height)
        : m_width(width), m_height(height), m_pixels(new Pixel(width * height))
        {
            init(0);
        }

        template<class Pixel, class SubPixel, int Layer>
        Image<Pixel, SubPixel, Layer>::~Image()
        {
            delete m_pixels;
        }


        template<class Pixel, class SubPixel, int Layer>
        Pixel& Image<Pixel, SubPixel, Layer>::operator() (int x, int y)
        {
            // implementation ...
        }

        template<class Pixel, class SubPixel, int Layer>
        const Pixel& Image<Pixel, SubPixel, Layer>::operator() (int x, int y) const
        {
            // implementation ...
        }

        template<class Pixel, class SubPixel, int Layer>
        SubPixel& Image<Pixel, SubPixel, Layer>::operator() (int x, int y, int subpixel)
        {
            // implementation ...
        }

        template<class Pixel, class SubPixel, int Layer>
        const SubPixel& Image<Pixel, SubPixel, Layer>::operator() (int x, int y, int subpixel) const
        {
            // implementation ...
        }

        template<class Pixel, class SubPixel, int Layer>
        void Image<Pixel, SubPixel, Layer>::init(Pixel emptyFill)
        {
            // implementation ...
        }

        template<class Pixel, class SubPixel, int Layer>
        inline bool Image<Pixel, SubPixel, Layer>::checkBounds(int x, int y, int subpixel) const
        {
            // implementation
        }

    }



}


// DepthImage.h:
namespace library
{
    namespace data
    {
        template class Image<unsigned short, unsigned short, 1>;
        class DepthImage : public Image<unsigned short, unsigned short, 1>
        {
            public:
            DepthImage(int width, int heigth);
            DepthImage(int width, int heigth, unsigned short emptyfill);
            ~DepthImage();

            protected:

            virtual unsigned short& pixelToSubpixel(unsigned short& pixel, int layer) const;
            private:
            DepthImage();

        };

    }
}

整个过程被编译成一个共享对象库。这个很好用。

一旦我尝试将一个可执行对象链接到这个.so-file，我就会收到以下消息：

library.so: undefined reference to `library::data::Image<unsigned short, unsigned short, 1>::Image(int, int, unsigned short)'
collect2: ld returned 1 exit status

我甚至想在DepthImage.h上做一个显式的模板实例化。由于DepthImage是而不是，是一个模板化的类(它只是在扩展一个类)，所以我不理解这个问题。

知道吗？？

你好，托拜厄斯

Answer 1

您没有为Image(int width, int height, Pixel emptyFill);提供定义，只为Image(int width, int height);提供定义。

Answer 2

错误消息中没有给出构造函数的定义。

Image(int, int, unsigned short)